How do you divide $\frac{x^{3} + 4 x^{2} + x - 6}{x - 3}$ $( x^3+4x^2+x-6 )/(x-3)$ ?

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Answer 1 · 2015-12-16T03:15:34

$\frac{(x^{2} + 1) (x + 4) (x - 6)}{x - 3}$ $((x^2+1)(x+4)(x-6))/(x-3)$

Assuming that you mean simplifying:

$\frac{x^{3} + 4 x^{2} + x - 6}{x - 3}$ $(x^3+4x^2+x-6)/(x-3)$

$= \frac{x^{2} (x + 4) + 1 (x - 6)}{x - 3}$ $=(x^2(x+4)+1(x-6))/(x-3)$

$= \frac{(x^{2} + 1) (x + 4) (x - 6)}{x - 3}$ $=((x^2+1)(x+4)(x-6))/(x-3)$ , $x \neq 3$ $x!=3$

How do you divide x3+4x2+x−6x−3( x^3+4x^2+x-6 )/(x-3)?